Graph Automorphism Group Equivariant Neural Networks

TL;DR

This work addresses learning from data with graph-structured symmetry by constructing neural networks that are equivariant to the graph automorphism group Aut(G). It provides a full characterization of learnable linear Aut(G)-equivariant maps between tensor powers of R^n via a spanning set of G-homomorphism matrices tied to $(k,l)$-bilabelled graphs, with a functorial link F^G between bilabelled graphs and linear maps. The main contributions include a formal framework that connects bilabelled-graph combinatorics to Aut(G)-equivariant layers, a spanning set of matrices X_H^G, and a pathway to recover the diagram basis for the symmetric group case through Frucht-inspired embeddings. The results enable principled design of Aut(G)-equivariant architectures for any finite group (via Frucht's theorem) and highlight both theoretical elegance and practical challenges, such as basis size growth and embedding-dependence. This lays groundwork for scalable, symmetry-aware GNNs and motivates future work on reducing the spanning set and efficient computation.

Abstract

Permutation equivariant neural networks are typically used to learn from data that lives on a graph. However, for any graph $G$ that has $n$ vertices, using the symmetric group $S_n$ as its group of symmetries does not take into account the relations that exist between the vertices. Given that the actual group of symmetries is the automorphism group Aut$(G)$, we show how to construct neural networks that are equivariant to Aut$(G)$ by obtaining a full characterisation of the learnable, linear, Aut$(G)$-equivariant functions between layers that are some tensor power of $\mathbb{R}^{n}$. In particular, we find a spanning set of matrices for these layer functions in the standard basis of $\mathbb{R}^{n}$. This result has important consequences for learning from data whose group of symmetries is a finite group because a theorem by Frucht (1938) showed that any finite group is isomorphic to the automorphism group of a graph.

Figures (6)

• Figure 1: The $(2,3)$--bilabelled graph diagram that is associated with the isomorphism class of $\bm{K} \coloneqq (K_3, (3,2), (3,3,1))$.
• Figure 2: The composition $[\bm{H_2}] \circ [\bm{H_1}]$ of the $(2,3)$--bilabelled graph diagram for $\bm{H_1} = (H_1, (3',2'), (1',2',3'))$ with the $(3,3)$--bilabelled graph diagram for $\bm{H_2} = (H_2, (2,1,2), (1,4,2))$, where $H_1$ is the graph having (relabelled) vertex set $[3']$ and edge set $\{(1',2')\}$ and $H_2$ is the graph having vertex set $[4]$ and edge set $\{(1,4)\}$. The vertices of the resulting bilabelled graph diagram on the RHS would be relabelled. For example, $\{1,2'\}$ could be relabelled as $1$ and $\{2,1',3'\}$ could be relabelled as $2$ to give the $(2,3)$--bilabelled graph diagram for $\bm{H} = (H, (2,1), (1,4,2))$, where $H$ is the graph having vertex set $[4]$ and edge set $\{(1,4), (1,2)\}$.
• Figure 3: For the graph $G$ having vertex set $V(G) = \{1, 2, 3\}$ and edge set $E(G) = \{(1,2)\}$, we show how to calculate the $(I,J)$-entries of the $G$-homomorphism matrix $X_{\bm{H}}^{G}$ corresponding to the isomorphism class $[\bm{H}]$ of the $(1,1)$--bilabelled graph $\bm{H} = (H, (3), (1))$ where $H$ is the graph having vertex set $V(H) = \{1, 2, 3, 4\}$ and edge set $E(H) = \{(1,2), (3,4)\}$. On the left hand side, we calculate the $(1,1)$-entry of $X_{\bm{H}}^{G}$. Since both of the black labelled vertices are being mapped to vertex $1$ in $G$, we can superimpose $1$ onto these vertices, and hence also onto the red labelled vertices that are connected with them, to determine where the other red vertices can be mapped to under a graph homomorphism. In this case, the only possible vertex in $G$ that the other red vertices can be mapped to is $2$. Hence there is only one possible graph homomorphism from $H$ to $G$ such that $1 \mapsto 1$ and $3 \mapsto 1$, and so the $(1,1)$-entry of $X_{\bm{H}}^{G}$ is $1$. On the right hand side, we calculate the $(3,2)$-entry of $X_{\bm{H}}^{G}$. We follow the same approach by superimposing $3$ in $G$ onto the black vertex labelled $1$ in $\bm{H}$, and $2$ in $G$ onto the black vertex labelled $3$ in $\bm{H}$. We relabel the red vertices that the black vertices are connected with and see where the other red vertices can be mapped to under a graph homomorphism. Whilst one of these red vertices can only be mapped to $1$ in $G$, the other red vertex cannot be mapped to any vertex in $G$, since the vertex $3$ in $G$ is not connected with any other vertex in $G$. Hence the number of graph homomorphisms from $H$ to $G$ such that $1 \mapsto 3$ and $3 \mapsto 2$ is $0$, and so the $(3,2)$-entry of $X_{\bm{H}}^{G}$ is $0$.
• Figure 4: We use Theorem \ref{['graphmainthrm']} to obtain the diagram basis of $\mathop{\mathrm{Hom}}\nolimits_{S_4}(\mathbb{R}^{4}, \mathbb{R}^{4})$ from all of the $(1,1)$--bilabelled graph diagrams.
• Figure 5: Depending on how $D_4$ is embedded as a subgroup of $S_4$, we obtain a basis of $\mathop{\mathrm{Hom}}\nolimits_{D_4}(\mathbb{R}^{4}, \mathbb{R}^{4})$, where here $D_4$ refers to the specific embedding in $S_4$ that is obtained from the different labellings of the vertices of $2K_2$, considered up to all automorphisms.

Theorems & Definitions (68)

• Definition 2.1
• Definition 2.2
• Definition 2.3
• Example 2.4
• Example 2.5
• Definition 2.6
• Definition 2.7
• Definition 2.8
• Remark 2.9
• Example 2.10